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If

is 45 degrees then only the body moves horizontal otherwise not.

Always consider moving from higer potential to lower potential as negative sign and moving from lower potential to higher potential as positive sign.

The same can be also considered reversed. But only one of them is to be considered.

(a) The current just after the connections are made is given by

i = (9 - 6) / 10 = 0.3 amp

(b) The current after completely charging is

i = (9 - 6) / 1 = 3 amp.

Nodal points method is same as Kirchoff's laws. Use Kirchoff's laws.

Method suggested by elessar_iitkgp is correct.

If the velocity of the ball when it strikes the incline is u, then after striking the incline its velocity remains same and makes an angle (alpha) w.r.t. the normal of the incline. Now use the equations for motion of a projectile on an incline plane.

Evaluation of the maximum ground acceleration distribution, seismic microzonation, is an essential procedure for prevention and mitigation of earthquake disasters.

The local acceleration effects, which are peculiar phenomena during atomic scale deposition process, were investigated byMolecular Dynamics (MD) simulation. The values of local acceleration were distributed widely for various surface orientations. Deposited atoms were accelerated along the potential energy surface, and accelerated values were evidently dependent on the local configuration of the surface. In contrast, the local acceleration became negligibly small for clusters consisting of many atoms.

Acceleration distribution is not neccessary for your syllabus.

This problem is from H.C.Verma.

Consider the first surface as plane and the first image is at ny, where n is the R.I. of the water and y is the distance from the plane surface at which the object is placed. This image works as object for the curved surface and the object distance from the second surface is (ny + h). Using the thick lens formula we get the second image distance v as

(1 / v) = (n - 1) / R - n / (ny + h) . Now replace v with -R. Because this image will work as object for the mirror and this object should appear to be at Center of Curvature. You will finaly get y as

y = (R - h) / n.

Wave equations are the same for your syllabus. That is you deal only with sinosoidal waves. These equations look like this.

Y = A sin ( kx + wt + q), This equation can be transverse or longitudinal. This can be for light or sound. We cannot make out until some details are given.

Since your syllabus does not deal with damping, you need not to worry.

Now (x - 2u) / (y - 2u) = 2 .

Solve for the requirement.

let the man velocity be v = u i + u j in vector form.
Let the wind velocity be w = x i + y j.

There fore the relative velocity = w - v = (x - u) i + (y - u) j
But the direction of the relative velocity is from north. Therefore x - u = 0. or x = u.

Now when the man velocity is doubled 2v = 2u i + 2u j.
therefore w - v = relative velocity.

substitute (EC - q) = u and you will find that dq = -du. rest is simple.

Dear debabratanag99,

I have already given you the answer, that scattering effect is for particles smaller in size than the wavelength. The wavelength of light is approximately from 3500 A to 7500 A. Theoretical models suggest that the average cluster may encompass as many as 90 H₂O molecules at 0°C, so that very cold water can be thought of as a collection of ever-changing ice-like structures. Therefore there is no point in discussing about scattering.

This explanation is flawed for more than one reason. Fog droplets are, on
average, smaller than cloud droplets, but they still are huge compared with
the wavelengths of visible light. Thus scattering of such light by fog is
essentially wavelength independent. Unfortunately, many people learn
(without caveats) Rayleigh's scattering law and then assume that it applies
to everything. They did not learn that this law is limited to scatterers
small compared with the wavelength and at wavelengths far from strong
absorption.

Dear MRIITIAN007,
You should know that I am an expert in this field. So before you could say anything wrong about my answer you should veryfy 100 times what you are going to say. I do not claim I am the best. Atleast for this standard of questions I am well experienced.
I rightly agree with you that you have freedom of expression. But it should be to express your answer not to coment about others aspecially the experts.
Good luck.
GoIIT team member.

Tyndall effect is not for FOG.